40æŠefghijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥¦§¨©ª«¬­®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÁÂÃÄÅÆÇÈÉÊËÌÍÎÏÐÑÒÓÔÕÖ×ØÙÚÛÜÝÞßàáâãäåæåæåæ„…†‡B–ú £kBhcIntegration by PartskBhc‚{Everything we say about antiderivatives comes from some derivative fact. Here is another derivative fact, the product rule: GrapQkBhcfGrapkBhcxkBhc  gGrapkBhcxopxEkBhcákBhc  = fopxEkBhcáGrapkBhcxkBhc  gGrapkBhcxkBhc  + fGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcf_If we rearrange the terms and integrate, this leads naturally to the following integral formulaetnIkBhcfopxEkBhcáGrapkBhcxkBhc  gGrapkBhcxkBhcxkBhc  = etnIGrapQkBhcfGrapkBhcxkBhc  gGrapkBhcxopxEkBhcákBhcxkBhc  - etnIkBhcfGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcxkBhcÉÂNotice the second integral. In this integral we are integrating a derivative - since integration and differentiation are inverses of each other, the integral and the derivative cancel to produceetnIkBhcfopxEkBhcáGrapkBhcxkBhc  gGrapkBhcxkBhcxkBhc  = fGrapkBhcxkBhc  gGrapkBhcxkBhc  - etnIkBhcfGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcx kBhc­¦The resulting formula is called the integration by parts formula. Its primary purpose is to transform difficult integrals to easier integrals (when properly applied). kBhcF?Here is how this can help you in practice. Consider the example etnI2kBhcx sin xkBhcx kBhcxThe integrand in this example is the product of two terms. We have to decide which of these two terms plays the role of kBhcfopxEkBhcáGrapkBhcxkBhc(! and which one plays the role of kBhcgGrapkBhcxkBhcXQ. The general rule of thumb that guides those choices is that one should pick an kBhcfopxEkBhcáGrapkBhcxkBhc-& term that is easy to integrate and a kBhcgGrapkBhcxkBhc˜‘ term with the characteristic that it gets simpler after differentiating. Given these requirements, the following are good choices for this case. kBhcfopxEkBhcáGrapkBhcxkBhc = sin xkBhcfGrapkBhcxkBhc = - cos xkBhcgGrapkBhcxkBhc  = xkBhcgopxEkBhcáGrapkBhcxkBhc  = 1kBhc' Putting this all together we get etnI2kBhcx sin xkBhcxkBhc  = etnIkBhcfopxEkBhcáGrapkBhcxkBhc  gGrapkBhcxkBhcxkBhc  = fGrapkBhcxkBhc  gGrapkBhcxkBhc  - etnIkBhcfGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcxkBhc = GrapkBhc- cos xkBhc  x - etnI2kBhc- cos xkBhcxkBhc= - x cos x + sin x + CkBhcHere is another exampleetnI1kBhc x ln xkBhcxkBhcrkSince the log term is hard to integrate, we are essentially forced into integrating the other term instead.kBhcfopxEkBhcáGrapkBhcxkBhc  = xkBhcfGrapkBhcxkBhc  = carF$kBhc1kBhc2kBhc  xopxEkBhc2kBhcxqThis is a good choice, because it now leaves us differentiating the log, which in fact produces a simpler result.kBhcgGrapkBhcxkBhc = ln xkBhcgopxEkBhcáGrapkBhcxkBhc  = carF$kBhc1kBhcxkBhc' Putting this all together yields etnITkBhcxopxEkBhc2kBhc  ln xkBhcxkBhc  = etnIkBhcfopxEkBhcáGrapkBhcxkBhc  gGrapkBhcxkBhcxkBhc  = fGrapkBhcxkBhc  gGrapkBhcxkBhc  - etnIkBhcfGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcxkBhc  = carF$kBhc1kBhc2kBhc  xopxEkBhc2kBhc ln x - etnI­carF$kBhc1kBhc2kBhc  xopxEkBhc2kBhc Grap0carF$kBhc1kBhcxkBhcxkBhc  = carF$kBhc1kBhc2kBhc  xopxEkBhc2kBhc ln x - carF$kBhc1kBhc2kBhc etnI,kBhcxkBhcx kBhc = carF$kBhc1kBhc2kBhc  xopxEkBhc2kBhc ln x - carF$kBhc1kBhc2GrapUcarF$kBhc1kBhc2kBhc  xopxEkBhc2kBhc  + C!kBhc:3Some more advanced examples of integration by parts#kBhcwpSometimes it is not immediately obvious that you can apply the integration by parts method. Consider the example'etnI3kBhcarctan xkBhcx(kBhc™’The problem here is that the integrand does not look like a product. However, a simple trick will make it a product: just introduce a factor of 1.)etnI5kBhc 1 arctan xkBhcx*kBhc*#This leads to the following choices+kBhcfopxEkBhcáGrapkBhcxkBhc  = 1,kBhcfGrapkBhcxkBhc  = x-kBhcgGrapkBhcxkBhc = arctan x.kBhcgopxEkBhcáGrapkBhcxkBhc  = carF@kBhc1kBhc 1 + xopxEkBhc2/kBhc(!Putting everything together gives0etnI3kBhcarctan xkBhcxkBhc  = fGrapkBhcxkBhc  gGrapkBhcxkBhc  - etnIkBhcfGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcx1kBhc= x arctan x - etnIqkBhc x carF@kBhc1kBhc 1 + xopxEkBhc2kBhcx2kBhcYRWe are almost there. The last thing we need to do is to introduce the substitution3kBhc u = 1 + xopxEkBhc24kBhcfor the last integral.5Dtot(kBhcukBhcxkBhc  = 2 x6 etnIqkBhc x carF@kBhc1kBhc 1 + xopxEkBhc2kBhcxkBhc  = carF$kBhc1kBhc2etnIscarF@kBhc1kBhc 1 + xopxEkBhc2kBhc  2 xkBhcxkBhc  = carF$kBhc1kBhc2etnIHcarF$kBhc1kBhcukBhcukBhc  = carF$kBhc1kBhc2kBhc ln u = carF$kBhc1kBhc2kBhc  lnGrap0kBhc 1 + xopxEkBhc27kBhcThe final result is8etnI3kBhcarctan xkBhcxkBhc = x arctan x - carF$kBhc1kBhc2kBhc  lnGrap0kBhc 1 + xopxEkBhc2kBhc  + C&kBhcB;Another famous example similar to the above is the integral%etnI/kBhc ln xkBhcx$kBhc The choices¥kBhcfopxEkBhcákBhc(x) = 1¦kBhcf(x) = x§kBhc g(x) = ln x¨kBhcgopxEkBhcákBhc (x) = carF$kBhc1kBhcx©kBhc make this problem doable:ªetnI/kBhc ln xkBhcxkBhc  = etnI1kBhc 1 ln xkBhcxkBhc = x ln x - x Grap0carF$kBhc1kBhcxkBhc = x ln x - 19kBhcReduction formulas:kBhcäÝMany integrals whose integrands look like some function to a power can be solved by some version of the method of integration by parts. Often, this method leads to what are known as reduction formulas. Here is an example.;etnITGrapkBhc sin xopxEkBhcnkBhcx<kBhc[TIn order to apply integration by parts, we have to write the integrand as a product.=etnIgGrapkBhc sin xopxEkBhc n-1kBhc  sin xkBhcx>kBhcb[Here are the assignments we need to make the integration by parts method work in this case.?kBhcfopxEkBhcáGrapkBhcxkBhc = sin x@kBhcfGrapkBhcxkBhc = -cos xAkBhcgGrapkBhcxkBhc  = GrapkBhc sin xopxEkBhc n-1B kBhcgopxEkBhcáGrapkBhcxkBhc  = GrapkBhc n-1kBhc GrapkBhc sin xopxEkBhc n-2kBhc  cos xCkBhc6/Substituting these assignments into the formulaDetnIkBhcfopxEkBhcáGrapkBhcxkBhc  gGrapkBhcxkBhcxkBhc  = fGrapkBhcxkBhc  gGrapkBhcxkBhc  - etnIkBhcfGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcxEkBhc givesFetnIgGrapkBhc sin xopxEkBhc n-1kBhc  sin xkBhcxkBhc = - cos x GrapkBhc sin xopxEkBhc n-1kBhc  - etnI“kBhc-cos x GrapkBhc n-1GrapkBhc sin xopxEkBhc n-2kBhc  cos xkBhcxGkBhc@9A simple transformation changes the integral on the rightHGrapkBhc n-1kBhc etnI–kBhc cosopxEkBhc2kBhc  x GrapkBhc sin xopxEkBhc n-2kBhc kBhcxIkBhc toJGrapkBhc n-1kBhc etnI±Grap?kBhc1 - sinopxEkBhc2kBhc  xkBhc GrapkBhc sin xopxEkBhc n-2kBhc kBhcxKkBhc#yielding these two integralsLGrapkBhc n-1etnIUkBhc sinopxEkBhc n-2kBhc  xkBhcxkBhc  - GrapkBhc n-1etnIFkBhc sinopxEkBhcnkBhcxMkBhcSubstituting this into feR F(1)kBhc  givesN etnIgGrapkBhc sin xopxEkBhc n-1kBhc  sin xkBhcxkBhc = - cos x GrapkBhc sin xopxEkBhc n-1kBhc  + GrapkBhc n-1etnIUkBhc sinopxEkBhc n-2kBhc  xkBhcxkBhc  - GrapkBhc n-1etnIFkBhc sinopxEkBhcnkBhcxOkBhc orP GrapkBhc n-1etnIFkBhc sinopxEkBhcnkBhcxkBhc  + etnIgGrapkBhc sin xopxEkBhc n-1kBhc  sin xkBhcxkBhc = - cos x GrapkBhc sin xopxEkBhc n-1kBhc + GrapkBhc n-1etnIUkBhc sinopxEkBhc n-2kBhc  xkBhcxQetnISkBhc sinopxEkBhcnkBhc  xkBhcxkBhc  = carF$kBhc1kBhcnGrapÑkBhc- cos x GrapkBhc sin xopxEkBhc n-1kBhc + GrapkBhc n-1etnIUkBhc sinopxEkBhc n-2kBhc  xkBhcxRkBhcçàThis is not really a complete answer, this is what we call a reduction formula. The idea with a reduction formula is to apply the formula repeatedly, reducing the exponent as we go along, until the exponent is either 1 or 0.SkBhcHere is an example.TetnISkBhc sinopxEkBhc5kBhc  xkBhcxkBhc  = carF$kBhc1kBhc5GrapµkBhc-cos x GrapkBhc sin xopxEkBhc4kBhc  + 4 etnISkBhc sinopxEkBhc3kBhc  xkBhcxUkBhcg`One application of the reduction formula is not enough, so we apply the reduction formula again.VetnISkBhc sinopxEkBhc3kBhc  xkBhcxkBhc  = carF$kBhc1kBhc3kBhc Grap’kBhc-cos x GrapkBhc sin xopxEkBhc2kBhc  + 2 etnI0kBhc sin xkBhcxWkBhcMFThe last integral we end up with is something we can do by inspection.XetnI0kBhc sin xkBhcxkBhc = -cos xYkBhcF?Substitution back in for the integrals we just computed, we getZetnISkBhc sinopxEkBhc3kBhc  xkBhcxkBhc  = carF$kBhc1kBhc3kBhc Grap{kBhc-cos x GrapkBhc sin xopxEkBhc2kBhc  + 2 GrapkBhc -cos x[etnISkBhc sinopxEkBhc5kBhc  xkBhcxkBhc  = carF$kBhc1kBhc5GrapkBhc-cos x GrapkBhc sin xopxEkBhc4kBhc  + 4 Grap»carF$kBhc1kBhc3kBhc Grap{kBhc-cos x GrapkBhc sin xopxEkBhc2kBhc  + 2 GrapkBhc -cos x\kBhcThis simplifies to] etnITGrapkBhc sin xopxEkBhc5kBhcxkBhc  = -carF%kBhc8kBhc 15kBhc cos x - carF%kBhc4kBhc 15kBhc cos x GrapkBhc sin xopxEkBhc2kBhc  - carF$kBhc1kBhc5kBhc cos x GrapkBhc sin xopxEkBhc4^kBhc"This same trick applies to _etnISkBhc cosopxEkBhcnkBhc  xkBhcx`kBhc(!and any number or other examples.akBhcAn alternative notationbkBhcibMany people use an alternative notation for the integration by parts formula that is more compact.cetnI,kBhcukBhcvkBhc = u v - etnI,kBhcvkBhcudkBhc®§The idea is still the same: we have an integral whose integrand is a product of two terms. We will end up integrating one of those terms and differentiating the other.ekBhcÍÆHere is one last example of integration by parts. This particular example is interesting because it demonstrates that sometimes integration by parts is useful even when it appears to not be working.fetnIUkBhceopxEkBhcxkBhc  cos xkBhcxgkBhcF?To apply integration by parts in this case, we make assignmentshkBhc u = eopxEkBhcxikBhcd u = eopxEkBhcxjkBhc d v = cos xkkBhc v = sin xlkBhcF?Putting these assignments into the integration by parts formulametnI,kBhcukBhcvkBhc = u v - etnI,kBhcvkBhcunkBhc givesoetnIUkBhceopxEkBhcxkBhc  cos xkBhcxkBhc  = eopxEkBhcxkBhc sin x - etnIJkBhcsin x eopxEkBhcxkBhcxpkBhc*#To handle the integral on the rightqetnIJkBhcsin x eopxEkBhcxkBhcxrkBhc/(we apply integration by parts again withskBhc u = eopxEkBhcxtkBhcd u = eopxEkBhcxukBhc d v = sin xvkBhc v = - cos xwkBhcwhich gives usxetnIJkBhcsin x eopxEkBhcxkBhcxkBhc  = eopxEkBhcxkBhc GrapkBhc- cos xkBhc  - etnIbGrapkBhc -cos xkBhc  eopxEkBhcxkBhcxykBhc Substituting feR x(3)kBhc back into feR o(2)kBhc  givesz etnIUkBhceopxEkBhcxkBhc  cos xkBhcxkBhc  = eopxEkBhcxkBhc sin x + eopxEkBhcxkBhc GrapkBhc  cos xkBhc  - etnIaGrapkBhc cos xkBhc  eopxEkBhcxkBhcx{kBhcèáAt first glance, we appear to have run in circles: we started with an integral and ended up transforming into the very same integral. Fortunately, some simple algebra will save the day. Adding the integral to both sides gives|kBhc 2 etnIJkBhccos x eopxEkBhcxkBhcxkBhc = cos x eopxEkBhcxkBhc  + eopxEkBhcxkBhc  sin x}kBhc or~etnIJkBhccos x eopxEkBhcxkBhcxkBhc  = carF$kBhc1kBhc2kBhc GrapjkBhccos x eopxEkBhcxkBhc  + eopxEkBhcxkBhc  sin xkBhc  + CkBhc which solves the problem.€kBhc2+Definite integrals and integration by partskBhcÝÖUp to this point, we have only used integration by parts to solve indefinite integrals (that is, integrals without limits). The method works just as well for definite integrations. Here is the formula in that case.‚ etnI™kBhcfopxEkBhcáGrapkBhcxkBhc  gGrapkBhcxkBhcxkBhcbkBhcakBhc  = fGrapkBhcxkBhc  gGrapkBhcxkBhc|sbuSkBhcaopxEkBhcbkBhc  - etnI™kBhcfGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcxkBhcbkBhcaƒkBhcIBHere is an example of an integration by parts problem with limits.ˆetnI‚kBhcxopxEkBhc2kBhc  eopxEkBhc -xkBhcxkBhc2kBhc1‰kBhcpiThe best way to proceed is to ignore the limits until we get to the very end of the problem. The integralŠetnIjkBhcxopxEkBhc2kBhc  eopxEkBhc -xkBhcx‹kBhc92suggests integration by parts with the assignmentsŒkBhc u = xopxEkBhc2kBhc d u = 2 xŽkBhcd v = eopxEkBhc -xkBhcv = - eopxEkBhc -xkBhc These assignments produce‘etnIjkBhcxopxEkBhc2kBhc  eopxEkBhc -xkBhcxkBhc  = xopxEkBhc2kBhc Grap/kBhc - eopxEkBhc -xkBhc  + etnITkBhceopxEkBhc -xkBhc  2 xkBhcx’kBhc The integral on the right“kBhc 2 etnIRkBhceopxEkBhc -xkBhc  xkBhcx”kBhc@9can be computed via integration by parts with assignments•kBhc u = x–kBhcd u = 1—kBhcd v = eopxEkBhc -x˜kBhcv = - eopxEkBhc -x™kBhc givingškBhc 2 etnIRkBhceopxEkBhc -xkBhc  xkBhcxkBhc  = 2GrapžkBhc x Grap.kBhc -eopxEkBhc -xkBhc  + etnIEkBhceopxEkBhc -xkBhcx›kBhcF?The final integral that appears is simple enough to do by hand.œetnIEkBhceopxEkBhc -xkBhcxkBhc = -carF<kBhc1kBhceopxEkBhcxkBhc  + CkBhc Substituting feR œ(6)kBhc back into feR š(5)kBhc  givesžkBhc 2 etnIRkBhceopxEkBhc -xkBhc  xkBhcxkBhc  = 2GrappkBhc x Grap.kBhc -eopxEkBhc -xkBhc  + -eopxEkBhc -xŸkBhc2+Substituting this result in turn back into feR ‘(4)kBhc  gives etnIjkBhcxopxEkBhc2kBhc  eopxEkBhc -xkBhcxkBhc  = xopxEkBhc2kBhc Grap/kBhc - eopxEkBhc -xkBhc  + 2GrappkBhc x Grap.kBhc -eopxEkBhc -xkBhc  + -eopxEkBhc -x¡kBhcd]Now that we have solved the problem without limits completely, we can re-introduce the limits¢etnI‚kBhcxopxEkBhc2kBhc  eopxEkBhc -xkBhcxkBhc2kBhc1kBhc  = GrapîkBhcxopxEkBhc2kBhc Grap/kBhc - eopxEkBhc -xkBhc  + 2GrappkBhc x Grap.kBhc -eopxEkBhc -xkBhc  + -eopxEkBhc -xkBhc|sbuSkBhc1opxEkBhc2£kBhc = Grap1kBhc -10 eopxEkBhc -2kBhc  - Grap0kBhc -5 eopxEkBhc -1